Advanced Calculus Single Variable

5.2 Exercises

Determine whether the following series converge and give reasons for your answers.

∑n=1∞

√n21+n+1-

∑n=1∞

(√n-+-1-− √n)

∑n=1∞

(n!)2
(2n)!

∑n=1∞

(2n)!
(n!)2

∑n=1∞

2n1+2

∑n=1∞

( )
nn+1

n

∑n=1∞

( )
-n-
n+1

n2

Determine whether the following series converge and give reasons for your
answers.

∑n=1∞

n
2n+2nn

∑n=1∞

2nn2+2nn

∑n=1∞

--n-
2n+1

∑n=1∞

-n100
1.01n

Find the exact values of the following infinite series if they converge.

∑k=3∞

k(1k−2)

∑k=1∞

--1--
k(k+1)

∑k=3∞

----1---
(k+1)(k−2)

∑k=1N

( )
1√k − √k1+1

Suppose ∑k=1∞ak converges and each ak≥ 0. Does it follow that ∑k=1∞ak2 also
converges?

Find a series which diverges using one test but converges using another if possible. If
this is not possible, tell why.

If ∑n=1∞an and ∑n=1∞bn both converge and an,bn are nonnegative, can you
conclude the sum, ∑n=1∞anbn converges?

If ∑n=1∞an converges and an≥ 0 for all n and bn is bounded, can you conclude
∑n=1∞anbn converges?

Determine the convergence of the series ∑n=1∞

(∑n 1)
k=1k

−n∕2.

Is it possible there could exist a decreasing sequence of positive numbers,

{an}

such that limn→∞an = 0 but ∑n=1∞

( )
1− an+1
an

converges? (This
seems to be a fairly difficult problem.)Hint: You might do something like this.
Show

-1−-x-- -1−-x--
lxim→1 − ln (x) = ln(1∕x) = 1

Next use a limit comparison test with

∞ ( )
∑ ln -an-
n=1 an+1

Go ahead and use what you learned in calculus about ln and any other techniques for
finding limits. These things will be discussed better later on, but you have seen them in
calculus so this is a little review.

Suppose ∑an converges conditionally and each an is real. Show it is possible to add the
series in some order such that the result converges to 13. Then show it is
possible to add the series in another order so that the result converges to
7. Thus there is no generalization of the commutative law for conditionally
convergent infinite series. Hint:To see how to proceed, consider Example
5.1.11.